Math Is Fun Forum

naturewild

Member

Registered: 2005-12-04

Posts: 30

Arrangement problem

1. Ten trees - four pines, four cedars, and two spruce- are planted in two parallel rows of five trees. How many arrangements are possible if each row must have the same composition of trees, not necessarily in the same order.

2. Suppose we want to creat subsets of the ten digits {0, 1, 2, ...., 9}
  a) How many subsets can be created, including the empty set?
  b) How many of the subsets contain only digits less than 7?
  c) How many of the subsets contain 0 or 9?

Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 48,395

Re: Arrangement problem

2 (a) 10C10+10C9+10C8+10C7+10C6+10C5+10C4+10C3+10C2+10C1+1(Empty set)
=1+10+45+120+210+252+210+120+45+10+1+1=1025.
(b)7C7+7C6+7C5+7C4+7C3+7C2+7C1+1 (Empty set)

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Pi Man

Guest

Re: Arrangement problem

1.   If the composite must be the same in each row, then each row contains 2 pines, 2 cedars and 1 spruce.   The first row can be arranged 30 different ways  - 5! / (2! * 2!).  For each of those 30 first rows, the second row could also be arranged 3o different ways.   So you have 30 * 30 = 900 different ways.

That's assuming you're treating all the pines the same, all the cedars the same and the 2 spruces the same.   If having spruce #1 and spruce #2 in the first row is to be treated differently then having spruce #3 and spruce #4 in that row, then the answer is a little more complicated.    For that, you need to consider how many different ways you can group the trees in addition to considering the order the trees are in.   You're still restricted to the 2 pines, 2 cedars and 1 spruce.   There are 6 ways to choose 2 pines (6 choose 2), 6 ways to choose 2 cedars and 2 ways to chose a spruce.   That means there are 72 different ways you can divide the trees into 2 rows.  For each of those 72 groups, you can shuffle the trees within the rows 900 different ways.   That makes 64800 different possibilities.

Pi Man

Guest

Re: Arrangement problem

There's a mistake in the 2nd part of my logic.   If it makes a difference which 2 spruces you have in row 1, then it should also make a difference where in the row each of those spruces are planted.   There are still 72 different groups but each of those groups could be arranged 120 differents ways.  Same with the 2nd row.   So the answer in that case would be 72 * 120 * 120 = 1036800.

Pi Man

Guest

Re: Arrangement problem

2c.)  It's probably easier to calculate how many subsets do NOT contain 0 or 9 and then subtract that from the total number of subsets.    Calculating the number of subsets containing only 1 through 8 is very similar to the way question 2b was calculated. The numbers of ways to use 1-8 only would be the same as calculating the number of subsets containing only digits less than 8.   Using Ganesh's short-hand:
= 8C8 + 8C7 + 8C6 + 8C5 + 8C4 + 8C3 + 8C2 + 8C1 + 8C0
=   1   +   8    +  28  +  56  + 70   + 56   + 28    + 8     +   1
=  256

Now, so that's the # of subnets containing 1-8, subtract that from the answer of 2a provided to us by Ganesh and we'll have the numbers of subsets containing 0 or 9.   1025 - 256 = 769

naturewild

Member

Registered: 2005-12-04

Posts: 30

Re: Arrangement problem

Thank you guys. The answer for no. 2 is all right.

However for the first one, the answer is 225. I'm not sure how.. I got 900 too.

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Arrangement problem

900 seems right to me, but 225 is 15 x 15, but if you do 180 degree rotation symmetry, then I think you might get 16 x 30, but it's just a guess.  I haven't looked thoroughly at it.

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